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Answer with explanation:</h2>
Formula to find the confidence interval for population proportion (p) is given by :-
, where z* = Critical value.
= Sample proportion.
SE= Standard error.
Let p be the true population proportionof U.S. adults who live with one or more chronic conditions.
As per given , we have
SE=0.012
By z-table , the critical value for 95% confidence interval : z* = 1.96
Now , a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions.:
Hence, a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions.
Interpretation : Pew Research Foundation can be 95% confident that the true population proportion (p) of U.S. adults who live with one or more chronic conditions lies between 0.42648 and 0.47352 .
The answer to the question
Answer:
The answer is either B or D.
Step-by-step explanation:
Associativ property
a(bc)=(ab)c
ability to move parenthaseese around
Answer:
Step-by-step explanation:
Given that sale of grills increase 6% per year
CUrrent sale of grills this year = 3300
So in the first year sales would increase by 3300(6%)
Total sales in the I year =3300+3300(6%)
This will be again increasing by 6% in II year and so on.
Hence we have
every year 3300 increases by 6% compoundly
So
No of sales in t year
=
In 6th year sale
s(6) =